Let $g(x)=\dfrac{10x}{x^3+5x}$ when $x\neq 0$. $g$ is continuous for all real numbers. Find $g(0)$. Choose 1 answer: Choose 1 answer: (Choice A) A $10$ (Choice B) B $2$ (Choice C) C $5$ (Choice D) D $0$
$\dfrac{10x}{x^3+5x}$ is continuous for all real numbers other than $x=0$ which means $g$ is continuous for all real numbers other than $x=0$. In order for $g$ to also be continuous at $x=0$, the following equality must hold: $\lim_{x\to 0}g(x)=g(0)$ We will obtain the above equality by letting $g(0)=\lim_{x\to 0}g(x)$. So let's find $\lim_{x\to 0}g(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to 0}g(x) \\\\ &=\lim_{x\to 0}\dfrac{10x}{x^3+5x} \gray{\text{This is the rule for }x\neq 0} \\\\ &=\lim_{x\to 0}\dfrac{10\cancel{(x)}}{\cancel{(x)}(x^2+5)} \gray{\text{Factor}} \\\\ &=\lim_{x\to 0}\dfrac{10}{(x^2+5)} \gray{\text{Cancel common factors}} \\\\ &\text{(This is allowed because }x\neq 0) \\\\ &=\dfrac{10}{(0)^2+5} \gray{\text{Direct substitution}} \\\\ &=2 \end{aligned}$ We obtained that if we set $g(0)=2$, then $\lim_{x\to 0}g(x)=g(0)$, which makes $g$ continuous at $x=0$. Since we already saw that $g$ is continuous for any other real number, we can determine that it's continuous for all real numbers. In conclusion, $g(0)=2$.